Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

The value of $\underset{x→α}{\lim}\frac{1-\cos(ax^2+bx+c)}{(x-α)^2}$, where $α$ and $β$ are the roots of $ax^2+bx+c=0$, is

Options:

$(a-b)^2$

$\frac{(α-β)^2}{2}$

$\frac{1}{2}a^2(α-β)^2$

none of these

Correct Answer:

$\frac{1}{2}a^2(α-β)^2$

Explanation:

$\underset{x→α}{\lim}\frac{1-\cos(ax^2+bx+c)}{(x-α)^2}$  $(\frac{0}{0}form)$

$=\underset{x→α}{\lim}\frac{(2ax+b)\sin(ax^2+bx+c)}{2(x-α)}=\underset{x→α}{\lim}\frac{(2ax+b)\sin[a(x-α)(x-β)]}{2(x-α)}$

$=\underset{x→α}{\lim}\frac{(2ax+b)\sin[a(x-α)(x-β)]}{2a(x-α)(x-β)}a(x-β)$

$=\frac{a^2}{2}\left(2α+\frac{b}{a}\right)(α-β)=\frac{a^2}{2}(2α-α-β)(α-β)=\frac{a^2(α-β)^2}{2}$